Archive for August 5th, 2009

I noticed some strange behavior in some Scala code recently. It was rather a mystery. I looked for my error and googled for a solution for the longest time with no success. Eventually I got my answer from the Scala mailing list / Nabble forum. Here’s the class that was causing the trouble.

The first thing you’ll notice about this class is that it is extremely simple! There aren’t a lot of moving parts. It’s a simple wrapper that exposes 3 basic array behaviors: apply (a ‘getter’), update (a ‘putter’), and good ol’ toString. Arrays in Scala take a type parameter, and to ensure that this class could wrap an array of any type I used a type parameter, too. Have a good look at the class and make sure you understand how it works. It won’t take long.

How do you expect this class to behave? Let’s play a little fill-in-the-blanks. Here is a Scala interpreter session with some results blanked out.

There are 6 blanks. What do you expect to see in each of those? Well, the first blank follows the creation of a new ArrayWrapper[Int] and its assignment to a val ‘a’. So, according to our overriding definition of toString, it is simply the result of the underlying Array’s toString method. I know from experience how a brand new Array of Ints looks. It looks like this:

scala> new Array[Int](5)
res1: Array[Int] = Array(0, 0, 0, 0, 0)

So that’s what I expect to see here. Anyone expect something different? Here’s what I actually saw in the first blank:

Hmm. That’s not what I expected. Did you predict this? Why is this array full of nulls when a new Array[Int] is usually full of zeros? I was stumped. The array is parameterized, I reasoned, so maybe type erasure was involved. That doesn’t make sense, though. No types should be erased at this point.

Let’s look at the next few lines, 12-16. I called a(0) (the apply method) and assigned the result to x. I then called the toString method on x. What do you expect in these two lines? I would have expected a(0) to return 0 and x.toString to return “0”, but my conviction is shaken by that last result. Will a(0) return null? Will x.toString throw a NullPointerException? Decide what you predict will happen. Here’s the actual result:

Each line behaves in the “correct” way even though we saw all those nulls in the underlying array. That’s good news, I suppose. Maybe the problem is limited to Array’s toString method. It should be smooth sailing now. Let’s now look at line 18, in which we call a(0).toString. It’s just combining the operations (apply and toString) from the previous two lines without storing the intermediate result in ‘x’. I expected that to return String “0”. You can probably guess by now that what I expected is not what I got. Make your own prediction before you read the actual result below. What will happen when we call a(0).toString?

scala> a(0).toString
java.lang.NullPointerException
at .<init>(<console>:7)
at .<clinit>(<console>)
at RequestResult$.<init>(<console>:3)
at RequestResult$.<clinit>(<console>)
at RequestResult$result(<console>)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at
sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
at sun.reflect.DelegatingMethodAccessorImpl.i...

Ouch! NullPointerException! This is an unpleasant surprise. The call a(0) returned a zero earlier, and calling toString on that zero returned a String “0”. But now we get this disaster. I’m getting more and more confused. Do you have an explanation for this crazy behavior yet?

Moving along, in line 21 we assign 0 to a(0). Remember that a(0) returned 0 earlier. By the way, behind the scenes the line “a(0) = 0” doesn’t call the apply method, but the ‘update’ method. It succeeds. In lines 23 and 26 we call a.toString and a(0).toString. What will happen in each case? At this point, it’s anybody’s guess. The behavior has been so wacky I can’t even make a sensible prediction. Make a guess of your own, if you dare, and observe the actual result below:

The underlying Array now appears to contain a zero in addition to the nulls. Also, the a(0).toString, which was throwing a NullPointerException earlier, is now succeeding.

As I say, I puzzled over this problem for some time. I wanted to blame the issue on type erasure in the parameterized type, but that explanation didn’t make sense. I posted a question to the Scala forum on Nabble and got a response back in short order from Daniel Sobral.

We declared ‘array’ to be an Array[A], which is to say an Array of who-knows-what. When the Array is defined in this way, with a type parameter of unknown type, the Array must be an array of object references! It cannot be an array of Java int primitives. That’s the peculiarity of parameterized types. That’s why the default values for the members of the array were null instead of 0. The underlying array is actually an array of java.lang.Integer objects.

When we ran ‘val x = a(0)’, Scala retrieved the value at index 0 which was null. The apply method has Int return type in our example, and null is not an legal value of an Int. Int is Scala’s version of the Java int primitive type. So the null was converted (unboxed) to Int value 0. Then it could be stored in val x, etc. Once it’s safely unboxed, it behaves like a normal Scala Int value.

So, why did a(0).toString not work? Shouldn’t the null returned from a(0) be unboxed to Int 0, then re-boxed for the toString call? Apparently it doesn’t work that way. The unboxing hasn’t happened at the time the toString call is executed, so that toString is called on the null, giving us the NullPointerException. I don’t know whether this behavior is imposed by the JVM or the Scala language. Either way, it seems to me like a violation of the Principle of Least Astonishment and an opportunity for improvement.

Once we call a(0) = 0, then the underlying array is populated with a boxed version of 0, which is to say an instance of java.lang.Integer. After it’s populated with a non-null it works normally.

Again, this only happens for Arrays with parameterized types. If we make ArrayWrapper non-parameterized and declare ‘array’ as an Array[Int] then the problem goes away.